Answer:
NH2.
Explanation:
The mass of hydrogen in the sample = 1.525 - 1.333 = 0.192g.
Dividing the 2 masses by the relative atomic mass of hydrogen and nitrogen:
H: 0.192 / 1.008 = 0.1905
N: 1.333 / 14.007 = 0.09517
The ratio of N to H = 0.09517 : 0.1905
= 1 : 2.
So the empirical formula is NH2.
4.79 x 10⁻⁷moldm⁻³
Explanation:
Given parameters:
pOH of RbOH = 6.32
Unknown:
Molarity of the base = ?
Solution:
The pH or pOH scale is used for expressing the level of acidity alkalinity of aqueous solutions.
pOH = -log[OH⁻]
we know the pOH to be 6.32
6.32 = -log[OH⁻]
[OH⁻] = inverse log₁₀(6.32)
[OH⁻] = 4.79 x 10⁻⁷moldm⁻³
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Answer:
2.50 atm
Explanation:
We have 10.4 g of DDT (solute), whose molar mass is 354.50 g/mol. The corresponding moles are:
10.4 g × (1 mol/354.50 g) = 0.0293 mol
The molarity of the solution is:
M = moles of solute / liters of solution
M = 0.0293 mol / 0.286 L
M = 0.102 M
We can find the osmotic pressure (π) using the following pressure.
π = M × R × T
where,
R: ideal gas constant
T: absolute temperature
π = M × R × T
π = 0.102 M × 0.0821 atm.L/mol.K × 298 K
π = 2.50 atm
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